2
$\begingroup$

I am looking for a reference for this result.

Let $S$ be a metric space such that

  • it is homeomorphic to a two-dimensional manifold,
  • it is 2-homogeneous: given two pairs of points $(x,y)$ and $(x',y')$ such that $d(x,y) = d(x',y')$ there exists an isometry of $S$ that sends $x$ to $x'$ and $y$ to $y'$.
  • it is a geodesic space: given two points $x$ and $y$, for $l=d(x,y)$ there exists a curve $\gamma:[0,l]\rightarrow S$ such that $d(\gamma(s), \gamma(t)) =|s-t|$, $\gamma(0) = x$, $\gamma(l) = y$.

Then $S$ is isometric to either the Euclidean plane, some hyperbolic plane, a sphere or its quotient by $\{\mathrm{id}, -\mathrm{id}\}$.

The result appears in a survey in French of Étienne Ghys in a volume entitled "L'héritage scientifique de Poincaré". There is no attribution, only a vague sketch of proof. I browsed Ghys' papers in search of this result without success.

$\endgroup$
6
  • 2
    $\begingroup$ I think it is in Busemann's work, either Geometry of Geodesics, or Metric methods in Finsler spaces and in the foundations of geometry $\endgroup$
    – Ben McKay
    Commented Sep 11, 2020 at 8:22
  • $\begingroup$ It's not true as stated: you should replace "isometric" with "homothetic". $\endgroup$
    – YCor
    Commented Sep 11, 2020 at 9:02
  • $\begingroup$ @YCor Indeed there are several hyperbolic and spherical spaces, classified up to isometry by the area of their unit disc. And only one euclidean space. So isometric to one of these. $\endgroup$
    – coudy
    Commented Sep 11, 2020 at 10:29
  • 1
    $\begingroup$ See theorem 7 in the survey arxiv.org/abs/1412.7893, and also arxiv.org/abs/2007.11917. $\endgroup$ Commented Sep 11, 2020 at 12:10
  • 3
    $\begingroup$ I mentioned this same theorem of Busemann’s at the end of an answer to another question, mathoverflow.net/a/346560. But the taxicab metric shows that Ghys’s first and third hypotheses on their own will not imply Busemann’s axiom for uniqueness of prolongation. $\endgroup$
    – user44143
    Commented Sep 12, 2020 at 6:22

0

You must log in to answer this question.